In the second paragraphs authors tries to prove ad absurdum that for weakly nuul sequence $\lim\limits_{n\to\infty}\Vert T(x_n)\Vert=0$. They say that without loss of generality one may suppose that $\{x_n\}_{n=1}^\infty$ is a weakly null sequence with $\Vert x_n\Vert=1$ and $\Vert T(x_n)\Vert>\delta$ for all $n\in\mathbb{N}$. I think they normalized original sequence $\{x_n\}_{n=1}^\infty$ and claims that it is also weakly null.

Why is weakly null sequence remains weakly null after normalization?

Another place I got stuck is the place where authors claims that passing to subsequence in $\{T(x_n)\}_{n=1}^\infty$ gives subsequence equivalent to the natural basis of $\ell_p$. And they also assume that after passing to subsequence $\{x_n\}_{n=1}^\infty$ remains to be equivalent to natural basis of $\ell_p$.

Why is $\{x_n\}_{n=1}^\infty$ remains to be equivalent to natural basis of $\ell_p$?

2 Answers
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We assume that $\{x_n\}$ doesn't converge strongly to $0$. We show, that each subsequence of $\{Tx_n\}$ has a convergent subsequence. So here, we can find a constant $C>0$ and $A$ infinite such that $\lVert x_n\rVert\geq C$ for all $n\in A$. Let $y_n:=\frac 1{\lVert x_n\rVert}x_n$ for $n\in A$. Then $\{y_n\}_{n\in A}$ converges weakly to $0$, as for $f\in (\ell^r)^*$ and $n\in A$, we have
$$|f(y_n)|=|f(x_n)|\frac 1{\lVert y_n\rVert}\leq \frac{|f(x_n)|}C.$$

By definition of equivalence $\{u_n\}$ and $\{v_n\}$ are equivalent if for all sequence $\{a_n\}$ of scalars, $\sum_{n=1}^{+\infty}a_nu_n$ is convergent if and only if so is $\sum_{n=1}^{+\infty}a_nv_n$. So $\{x_n\}_{n\in A}$ is equivalent to $\{e_n\}_{n\in A}$ (not to the whole sequence). But it's enough to conclude, as we would have boundedness from an infinite subspace of $\ell^r$ to $\ell^p$.

If $\|x_{n}\|\rightarrow 0$, then $x_{n}\rightarrow 0$ which implies that $T(x_n)\rightarrow 0$. So you can suppose that $\|x_{n}\|>\delta$ for some $\delta>0$. By hypothesis you have $$\langle y,x_{n}\rangle\rightarrow 0,\ \forall\ y\in X^{\star}$$